﻿ Rational Functions Word Problems | Problems & Solutions

# Rational Functions Word Problems

Rational Functions Word Problems
• Page 1
1.
What is the solution of the equation + = 2?
 a. 1 and 5 b. - 1 and 7 c. - 1 and - 5 d. 1 and - 5

#### Solution:

2x Ã¢â‚¬â€œ 2 + 32x2 Ã¢â‚¬â€œ 5x + 6 = 2
[Original equation.]

2x - 2 + 32(x - 2)(x - 3) = 2
[Factor denominator.]

The LCD is (x - 2)(x Ã¢â‚¬â€œ 3).

2x - 2(x - 2)(x - 3) + 32(x - 2)(x - 3)(x - 2)(x - 3) = 2 · (x Ã¢â‚¬â€œ 2)(x Ã¢â‚¬â€œ 3)
[Multiply each side by LCD.]

2(x - 3) + 32 = 2(x - 2)(x - 3)
[Simplify.]

2x2 Ã¢â‚¬â€œ 12x - 14 = 0
[Write in standard form.]

(x + 1)(2x Ã¢â‚¬â€œ 14) = 0
[Factor.]

x = - 1 and x = 7
[Solve.]

The solutions are - 1 and 7.

2.
Ethan takes 15 minutes and Jake takes 60 minutes for the newspaper distribution in Rosedale. How long would it take for both of them, working together, to distribute the newspaper in Rosedale?
 a. 15 minutes b. 12 minutes c. 12 minutes 30 seconds d. 15 minutes 20 seconds

#### Solution:

Part of distribution done by Ethan in 1 minute = 1 / 15

Part of distribution done by Jake in 1 minute = 1 / 60

Let t be the time spent by Ethan and Jake in distributing the newspaper.

[(Part of distribution done by Ethan) × (Time taken by Ethan)] + [(Part of distribution done by Jake) × (Time taken by Jake)] = 1

(1 / 15)t + (1 / 60)t = 1
[Writing algebraic model.]

[(1 / 15)t × 60] + [(1 / 60)t × 60] = 1 × 60
[Multiply both sides with the LCD 60.]

4t + t = 60
[Simplify.]

5t = 60
[Combining like terms.]

t = 60 / 5 = 12
[Dividing each side by 5.]

Ethan and Jake together will distribute the newspaper in Rosedale in 12 minutes.

3.
Solve:
$\frac{2}{x-3}$ = $\frac{x}{2}$
 a. - 4 and - 1 b. 4 and - 1 c. - 4 and 1 d. 4 and 1

#### Solution:

2x-3 = x2
[Original equation.]

(2 × 2) = [x × (x - 3)]
[Cross multiply.]

4 = x2 - 3x
[Simplify each side of the equation.]

x2 - 3x - 4 = 0
[Write the equation in the standard form.]

(x - 4)(x + 1) = 0
[Factor.]

x = 4 and x = - 1
[Solve.]

The solutions of the equation are 4 and - 1.

4.
Solve:
$\frac{x}{15}$ = $\frac{9}{5}$
 a. 135 b. 28 c. 27 d. $\frac{9}{5}$

#### Solution:

x15 = 95
[Original equation.]

5x = 135
[Cross multiply.]

x = 27
[Divide each side by 3.]

The solution of the equation is 27.

5.
Annie can do a job in 4 days and her assistant, Andrew can do the same job in 16 days. How long will it take for Andrew and Annie to complete the job, if they work together?
 a. 2 days b. $3\frac{1}{5}$ days c. $3\frac{4}{5}$ days d. $3\frac{3}{5}$ days

#### Solution:

Part of the job Annie can do in 1 day = 1 / 4

Part of the job Andrew can do in 1 day = 1 / 16

Let t be the time taken by Annie and Andrew to complete the job.

[(Part of the job Annie can do in a day) × (Time taken by Annie)] + [(Part of the job Andrew can do in a day) × (Time taken by Andrew)] = 1

(1 / 4)t + (1 / 16)t = 1
[Write an algebraic model.]

[16 × (1 / 4)t] + [16 × (1 / 16)t] = 16 × 1
[Multiply both sides with the LCD 16.]

4t + t = 16
[Simplify.]

5t = 16
[Combine like terms.]

t = 16 / 5 = 31 / 5
[Divide each side by 5.]

The time taken by Annie and Andrew to complete the job together is 31 / 5 days.

6.
Solve:
=
 a. ± 4 b. ± 7 c. ± 6 d. ± 5

#### Solution:

4x + 3 = x - 34
[Original equation.]

(4 × 4) = (x + 3) × (x - 3)
[Cross multiply.]

16 = x2 - 9
[Simplify.]

25 = x2

x = ± 5
[Find the square roots].

The solutions of the equation are + 5 and - 5.

7.
Solve:
= $\frac{1}{4}$
 a. - 21 b. 19 c. - 5 d. 22

#### Solution:

5(s2 + 1)(20s2 - s - 1) = 1 / 4
[Original equation.]

4 × 5(s2 + 1) = 1 × (20s2 - s - 1)
[Cross multiply.]

20s2 + 20 = (20s2 - s - 1)
[Simplify each side of the equation.]

20 = - s - 1
[Subtract 20s2 from each side.]

s = - 20 - 1 = - 21
[Simplify.]

The solution is - 21.

8.
Solve:
$\frac{3}{\left(2p-1\right)}$ + $\frac{2}{\left(2p+1\right)}$ = 0
 a. - 1 b. $\frac{1}{10}$ c. - $\frac{1}{10}$ d. 10

#### Solution:

3(2p-1) + 2(2p+1) = 0
[Original equation.]

3(2p-1) = - 2(2p+1)
[Subtract 2(2p+1) from each side.]

[3 × (2p + 1)] = - [2 × (2p - 1)]
[Cross multiply.]

6p + 3 = - 4p + 2
[Simplify each side of the equation.]

10p + 3 = 2

10p = - 1
[Subtract 3 from each side].

p = - 1 / 10
[Divide each side by 10.]

The solution of the equation is - 1 / 10.

9.
Solve:
$\frac{4}{s}$ - $\frac{1}{4s}$ = $\frac{3}{4}$
 a. 5 b. - 7 c. 7 d. - 5

#### Solution:

The LCD of s and 4s is 4s.

4s - 14s = 3 / 4
[Original equation.]

(4s × 4s) - (4s × 14s) = (4s × 3 / 4)
[Multiply each side by the LCD, 4s.]

16 - 1 = 3s
[Simplify.]

5 = s
[Solve.]

The solution of the equation is 5.

10.
Solve:
$\frac{5}{\left(x+2\right)}$ = ($x$ - 2)
 a. 2 and - 2 b. 4 and - 4 c. 5 and - 5 d. 3 and - 3

#### Solution:

5(x+2) = (x - 2)
[Original equation.]

5 = (x - 2) × (x + 2)
[Cross multiply.]

5 = x2 - 4
[Simplify the right-hand side of the equation.]

9 = x2